The New Sudoku Players' Forum

[dgwil]

The beginning numbers are

Code: Select all

..1..8...
.3..7..9.
9.....6..
..4..3..2
.2.....4.
7..1..5..
..8.....1
.6..2..8.
...6..7..


My wits end is below. This may or may not require something advanced but to get this far I’ve used the basic hidden and naked pairs and triples, and other elimination basic stuff. I sort of understand the examples of some advanced type things, x and y wings, uniqueness etc, but have problems spotting them, even after unusually long periods of staring. Please help direct me to the next moves and the reasons behind them. Thanks in advance, Wil

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------------+------------+-----------+
|25  47   1 | 9   6   8  |234 257 345|
|8   3    6 | 2   7   45 |1     9  45|
|9   47  25 | 3   1   45 |6   257   8|
------------+------------+-----------+
|6   5   4  | 7   9   3  |8    1   2 |
|1   2   3  | 8   5   6  |9    4   7 |
|7   8   9  | 1   4   2  |5    3   6 |
------------+------------+-----------+
|245 9   8  | 45  3   7  |24   6   1 |
|345 6   7  | 45  2   1  |34   8   9 |
|2345 1  25 | 6   8   9  | 7   25 345|


[Wolfgang]

Look at the 5 in row 9 and box 9

[dgwil]

Thanks Wolfgang, The 5s in box 9 can only go on row 9 eliminating the 5's in box 7 column 1&3. Leaves me a lone 2 in column 3 which will probably be enough to solve this sucker. That's one elimination procedure i'm supposed to know, just didn't stare long enough. Thanks again, Wil

[dgwil]

Dang wolfgang, looking at it again there an obvious naked pair (25) column 3 & 8 that also eliminates 5s row 9. I guess I stared at this thing so long my eyes glazed over when I reached row 9.Thanks again for the heads up, Wil

[ronk]

The 5's in box 9 can only go on row 9 eliminating the 5's in box 7 column 1&3 ...

edit: AND later ...

an obvious naked pair (25) column 3 & 8 that also eliminates 5's row 9

If you notice both of those simultaneusly, i.e., before the consequences of one destroy the other, then you might proceed directly to placing the 5 in row 9 col 8.

[Cec]

"..The 5's in box 9 can only go on row 9 eliminating the 5's in box 7 column 1&3. .."

Hi dgwil,
Sorry but not quite right. This particular elimination technique is known as "Locked Candidates(1) and is one of a number of techniques described by clicking here. In row 9 the 5's "Locked" in box 9 means any other 5's can be safely excluded only from this same row 9 (outside of box9) but not all the candidate 5's in box7.

Cec

[ronk]

"..The 5's in box 9 can only go on row 9 eliminating the 5's in box 7 column 1&3. .."

Hi dgwil,
Sorry but not quite right. ... In row 9 the 5's "Locked" in box 9 means any other 5's can be safely excluded only from this same row 9 (outside of box9) but not all the candidate 5's in box7.

Cec

Even if not so expressed, I believe that's what was meant.

[dgwil]

Hello cvcbevwr,
You’re correct, I knew but stated incorrectly, “eliminating the 5s in box 7” meaning 5s on row 9 outside of box 9. However, thanks for making me hip to these 5s being called “locked candidates”. I was just startled, after Wolfgang suggested I look at the 5s in r9box9, to see a couple of elimination type things that I know how to do, jump off the page at me, after I had starred at the thing for over and hour. I only know how to do a few things and naked pairs and locked candidates are among them. On another note, I try, from the way I understand uniqueness mentioned on this board, to use it to eliminate on some uniqueness situations. EDIT, THIS IS IF THE PUZZEL WERE NOT SOLVED WHICH IT HAS BEEN, END EDIT It would not advance me very far but look at column 7. Could I eliminate the 4 at r1c7due to the 2 & 3 at r7&8c7 leaving me a 45 at r7&8c1. This would leave, I believe, a to be avoided, doomsday, multible solution situation with the 45 at r7&8c1 and 45 at r7&8c4. Thanks for any input, Wil[/b]

[ronk]

It would not advance me very far but look at column 7. Could I eliminate the 4 at r1c7due to the 2 & 3 at r7&8c7 leaving me a 45 at r7&8c1. This would leave, I believe, a to be avoided, doomsday, multible solution situation with the 45 at r7&8c1 and 45 at r7&8c4.

That is a valid elimination, but allow me to restate why it is valid.

Code: Select all

------------+------------+-----------+
|25  47   1 | 9   6   8  |234 257 345|
|8   3    6 | 2   7   45 |1     9  45|
|9   47  25 | 3   1   45 |6   257   8|
------------+------------+-----------+
|6   5   4  | 7   9   3  |8    1   2 |
|1   2   3  | 8   5   6  |9    4   7 |
|7   8   9  | 1   4   2  |5    3   6 |
------------+------------+-----------+
|245 9   8  | 45  3   7  |24   6   1 |
|345 6   7  | 45  2   1  |34   8   9 |
|2345 1  25 | 6   8   9  | 7   25 345|

Because of the "unique rectangle" at r78c14, both eliminations r7c1<>2 and r8c1<>3 may not occur for a puzzle with a unique solution. In other words:

If r7c1<>2, then r8c1=3 and if r8c1<>3, then r7c1=2. Using these implications, there exists a "closed loop" such that ...

... r7c1<>2 implies r8c1=3 implies r8c7=4 implies r7c7=2 implies r7c1<>2 ... (now looping)

and going the opposite direction ...

... r8c1<>3 implies r7c1=2 implies r7c7=4 implies r8c7=3 implies r8c1<>3 ...

In either case, a 4 exists in the intersection of col 7 and box 9. Therefore, all other 4s in those units may be safely eliminated ... meaning r1c7<>4 and r9c9<>4.

Ron

[dgwil]

Thanks Ron,
That helps me to see the process behind some of these principles. Also, as you are pointing out, I spoke to soon when I said, “It would not advance me very far”. Eliminating the 4 at r1c7 gives me the locked candidate 4 in box 3 giving the r9c9<>4 that you explained. This forces the 4 to r1c9, leaving the naked pair 25 at r1c1 and r7c1, forcing r8c1=3,r8c7=4,r7c7=2, r7c1=5 , r1c1=2 and r9c3=2 eventually solving the grid. As you pointed these same solutions come from the proper way you applied the uniqueness principle and the thought behind it. Had the original 5s on row 9 not have been so easy and obvious then uniqueness would be a lifesaver to proceed with the eliminations of the puzzle.
Thanks again for your input, Wil